The number of solutions to `x+y+z=10`, where `1le x, y, z le 6` and `x, y, z in N`, is equal to

Options

A. 35
B. 36
C. 27
D. 66

Answer

To find the number of solutions to the equation \( x + y + z = 10 \) where \( 1 \leq x, y, z \leq 6 \) and \( x, y, z \in \mathbb{N} \), we can follow these steps:

### Step 1: Change of Variables
Since \( x, y, z \) must be at least 1, we can substitute:
- Let \( x = a + 1 \)
- Let \( y = b + 1 \)
- Let \( z = c + 1 \)

This means \( a, b, c \) must be non-negative integers (i.e., \( a, b, c \geq 0 \)).

### Step 2: Substitute in the Equation
Substituting these into the original equation gives:
\[ (a + 1) + (b + 1) + (c + 1) = 10 \]
This simplifies to:
\[ a + b + c + 3 = 10 \]
Thus, we have:
\[ a + b + c = 7 \]

### Step 3: Count the Non-negative Integer Solutions
We need to find the number of non-negative integer solutions to the equation \( a + b + c = 7 \). The formula for the number of solutions in non-negative integers to the equation \( x_1 + x_2 + \ldots + x_r = n \) is given by:
\[ \binom{n + r - 1}{r - 1} \]
where \( n \) is the total sum and \( r \) is the number of variables.

In our case, \( n = 7 \) and \( r = 3 \):
\[ \text{Number of solutions} = \binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} \]

### Step 4: Calculate the Binomial Coefficient
Calculating \( \binom{9}{2} \):
\[ \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = \frac{72}{2} = 36 \]

### Step 5: Adjust for Upper Limits
Now, we need to ensure that \( x, y, z \) do not exceed 6. Since we replaced \( x, y, z \) with \( a + 1, b + 1, c + 1 \), the upper limit translates to \( a, b, c \leq 5 \).

### Step 6: Subtract Invalid Solutions
We need to find the cases where one of \( x, y, z > 6 \). Suppose \( x > 6 \), then \( x \geq 7 \) implies \( a + 1 \geq 7 \) or \( a \geq 6 \). Let \( a' = a - 6 \) (where \( a' \geq 0 \)):
\[ a' + b + c = 1 \]
The number of non-negative integer solutions to this equation is:
\[ \binom{1 + 3 - 1}{3 - 1} = \binom{3}{2} = 3 \]
This is the same for \( y > 6 \) and \( z > 6 \), so we subtract \( 3 \times 3 = 9 \) from our previous total.

### Step 7: Final Count of Solutions
Thus, the total valid solutions are:
\[ 36 - 9 = 27 \]

### Conclusion
The number of solutions to the equation \( x + y + z = 10 \) under the given constraints is \( \boxed{27} \).

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